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Constitutional supercooling occurs when liquid ahead of a solidification phase boundary lies below its local liquidus. In a saline liquid with $T_L=-mC$ and $m>0$, salt rejection enriches the interface liquid and depresses its liquidus; as the concentration decreases away from the interface, the liquidus rises. If it rises faster than the actual temperature, a supercooled interval appears. This can make a planar solidification phase boundary unstable.

In a constant-thickness laminar boundary layer at an ablating glacier, approximate fluxes by $\kappa (T_{\infty }-T_i)/\delta$ and $D(C_{\infty }-C_i)/\delta$. Neglect solid diffusion and sensible heating relative to latent heat. With ablation speed $U>0$, solute balance gives $U(C_i-C_s)=D(C_{\infty }-C_i)/\delta$, so $C_i=C_s+(C_{\infty }-C_s)/(1+P)$, $P=U\delta /D$. The liquidus $T_i=T_m-\Gamma (C_i-C_s)$ and Stefan condition $UL/c_p=\kappa (T_{\infty }-T_i)/\delta$ then give the displayed law, where $A_T=LD/(c_p\kappa )$. Its positive root exists when the ocean is above its own freezing point. In the small-$P$ regime, interface salinity approaches ocean salinity; in the large-$P$ regime it approaches the ice composition.

For a linear imposed temperature $T=T_E+Gz$, local liquidus equilibrium gives $C=-(T_E+Gz)/m$ in the mushy layer. With equal densities, pure solid, pulling velocity $-V$, and diffusion only in liquid volume fraction $q=1-\varphi$, steady solute flux is $-VqC-Dq{C}_z=-V{C}_0$. The exterior concentration is $C_0+(C_i-C_0)e^{-V(z-h_0)/D}$. Marginal equilibrium at a mush-liquid boundary gives $C_i-C_0=GD/(mV)$ and hence the displayed thickness. In the mush, $q[C+(D/V)C_z]=C_0$, so $\varphi =(h_0-z)/(m{C}_0/G+h_0-z)$. A positive thickness requires the thermal interval to exceed the diffusive length $D/V$.

For pure solid rejecting all solute, pulling speed $V>0$, solutal diffusivity $D$ and finite solute inventory per unit area $N$, a stationary planar front has $C(z)=C_I{e}^{-Vz/D}$ in its liquid half-space. The advection-diffusion equation gives this profile, and ${\int }_0^{\infty }Cdz=N$ fixes $C_I=VN/D$. A liquidus $T_I=-m{C}_I$ fixes its position in an imposed temperature gradient. The pure-solid branch requires $C_I<C_E$, where $C_E$ is the eutectic composition.

Freezing is a liquid-to-solid phase transition. A moving phase boundary releases latent heat; its rate is constrained by heat transport and the Stefan condition. In a solution the liquidus depends on composition, so being below the pure solvent's melting temperature does not imply that the solution must freeze.

The frozen-temperature approximation prescribes the external thermal field during interface perturbations, rather than solving a perturbation heat equation. An interface displacement samples that prescribed field at its new location. Local liquidus and Gibbs-Thomson effect conditions still apply, and the solute field may retain full time-dependent diffusion.

Salt lowers the freezing temperature of water. Consequently ice can ablate in saline water below the pure-water melting temperature if the ocean is warmer than its own liquidus. Solute diffusion towards the interface sustains the freezing-point depression. In the diffusive glacier-ablation interface law, small compositional Péclet number makes interfacial salinity close to ocean salinity and gives $P\simeq [T_{\infty }-T_m+\Gamma (C_{\infty }-C_s)]/[LD/(c_p\kappa )+\Gamma (C_{\infty }-C_s)]$ near the no-ablation threshold. This is diffusion-assisted dissolution rather than a supply of heat above the pure-water melting point; latent heat must still be supplied to complete the phase change.

In a planar no-flow model with pure ice, a stable cold brine below freshwater can support an ice layer whose upper phase boundary freezes freshwater while its lower boundary melts into diluted brine. Heat conduction and salt diffusion obey different similarity scales. Local liquidus equilibrium, salt conservation and two Stefan conditions determine both boundary positions. The heat lost into the brine supports net ice growth even while some basal ice melts.

Instantaneous interface kinetics is the limit of negligible kinetic undercooling at a moving solidification interface. Its temperature is set by local liquidus equilibrium and curvature-induced melting-temperature depression, without a further velocity-dependent attachment term. It does not imply that diffusion or heat transport is instantaneous.

An ice bath can absorb heat while its temperature barely changes: the heat converts ice into water. The latent heat of fusion quantifies this melting budget. For a pure substance at its equilibrium melting temperature $T_m(p)$ and specified pressure $p$, define it here per unit mass:Here $h_l$ and $h_s$ are the specific enthalpies of the coexisting bulk liquid and solid, and $H$ is enthalpy. For ordinary melting, $L>0$, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$; an amount $m$ has a total fusion enthalpy $mL$ in joules. A molar fusion enthalpy instead has units $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$ and equals $ML$, where $M=m/n$ is mass divided by amount of substance in $\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$. Always check the normalization before inserting a tabulated value into an energy balance.^{[ref] [ref]}

For a closed system converting completely from solid to liquid at fixed pressure, with only pressure-volume work and negligible changes in bulk kinetic energy and potential energy, the first law of thermodynamics gives $Q_p=\Delta H=mL$. Both endpoints are at $T_m(p)$: warming initially colder ice or subsequently warming the resulting water requires additional sensible heat. Since $H=U+pV$, the corresponding specific internal energy jump is $\Delta u=L-p\Delta v$, where $v=V/m$; $L$ includes the work associated with the change in volume. Heat is transferred energy, whereas enthalpy is a state function. ^[ref]

Let $x=m_l/m$ be the liquid phase fraction by mass, rather than by volume. Neglecting interfacial contributions to enthalpy, equilibrium phase coexistence at fixed pressure givesIn the model above, reaching $T_m$ takes $2.10\mathrm{kJ}$ of sensible heat; the plateau then absorbs $33.40\mathrm{kJ}$, ending at a cumulative $35.50\mathrm{kJ}$. At $18.80\mathrm{kJ}$, half the mass is liquid and the temperature is still $T_m$. The final warming takes another $4.18\mathrm{kJ}$. The rounded $L$ and $c_l$ are inspired by the introductory ice-melting calorimetry values in Case Western's demonstration; constant properties and exact equilibrium throughout are stipulated model assumptions. ^[ref]

Within one thermodynamic phase, specific heat capacity at constant pressure is $c_p=(\partial h/\partial T{)}_p$, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$; a sample's heat capacity at constant pressure is $C_p=m{c}_p$, in $\mathrm{J}{\mathrm{K}}^{-1}$. The plateau is a finite enthalpy change at zero temperature change through varying $x$, so neither a finite single-phase $c_p$ nor $Q=m{c}_p\Delta T$ represents it. The heating curve needs separate sensible heat intervals and a fusion contribution. ^[ref]

The plateau assumes a pure substance, a stable bulk solid-liquid phase coexistence pair, fixed pressure, and sufficiently slow, uniform heating to remain near thermodynamic equilibrium. The total mass and composition must remain fixed, with negligible evaporation and no other phase transition. A fast heater can produce temperature gradients, so a local thermometer need not read the equilibrium $T_m$. Neither $T_m$ nor $L$ is universal across pressures or compositions: common-pressure melting-point slope describes the pressure dependence of $T_m$, while mixtures can melt over a temperature range as their coexisting compositions change. Their liquidus must be consulted instead of assuming a pure-substance plateau. ^[ref]

This makes ice useful as a temperature buffer: under these conditions it can absorb $m_{s}L$ before the remaining solid mass $m_s$ disappears, after which the water warms. If a stipulated constant net input power $P$ goes entirely into melting, that buffer lasts $m_{s}L/P$. Reversing the equilibrium conversion releases the same fusion enthalpy during freezing. The beautiful thermodynamic connection is $L=T_m(s_l-s_s)$, with $s$ the specific entropy and $T_m$ absolute temperature: equal Gibbs free energies at phase coexistence allow different enthalpies and entropies. The nearly stationary thermometer hides a finite, measurable change of state. ^[ref]

At a sharp mush–liquid edge with linear liquidus $T_L=T_m-mC$, marginal equilibrium makes the liquid temperature gradient tangent to the local liquidus: $T_z=-m{C}_z$. Together with $T=T_L$ at the edge and $T\ge T_L$ in the exterior liquid, it supplies the usual absence-of-constitutional supercooling closure. When solute diffusion is neglected inside the mushy layer but retained outside it, interfacial solute conservation can require a small solid fraction jump; imposing unit porosity as an additional finite-diffusivity condition generally overdetermines this sharp-edge reduction.

Melting is a solid-to-liquid phase transition that absorbs latent heat. It can be driven by heating, changing phase pressures, or adding solute that causes freezing-point depression. The ice layer between freshwater and cold brine illustrates melting into a cold liquid through dilution and a composition-dependent liquidus.

A mushy layer contains solid crystals and liquid in local phase equilibrium, rather than a single sharp phase boundary. Its solid fraction changes as heat and solute are transported. A concentration-dependent liquidus couples temperature to liquid composition; a connected crystal framework makes the residual liquid behave as a porous-media flow.

For nonzero imposed thermal gradient and linear liquidus equilibrium, $C_z=-G/m\ne 0$ wherever a mixture of crystals and pore liquid persists. A sealed boundary requires zero solute flux, so a local-equilibrium mush trace there would require $1-\varphi =0$, not a zero-solid-fraction roof. In particular a pure-liquid trace with the equilibrium concentration gradient has nonzero diffusive flux. A sealed column must develop a nonequilibrium-in-the-liquid temperature/composition relation, a separate liquid cap, or an appropriate phase-fraction jump; simply transporting the initial zero-fraction level violates the boundary condition.

Write $H=L/c_p$, $k=\rho c_p\kappa$ and $ϵ=\sqrt{D/\kappa }$. The requested model neglects density differences and uses the same thermal transport coefficients in the interfacial balances. Let $C_b=C(b(t),t)$ and $T_b=T_m-m{C}_b$. With no imposed heat gradient in the fresh liquid, its temperature remains $T_m$. The prescribed linear temperature in the ice isThe salt and temperature fields in the lower liquid satisfy diffusion equations, $C_t=D{C}_{zz}$ and $T_t=\kappa T_{zz}$. For constant interfacial values and the proposed similarity solution, their decaying solutions areThese follow by substituting a variable $z/(2\sqrt{\chi t})$ into the diffusion equation, which gives $U^{\prime \prime }+2\eta U^{\prime }=0$; integrating gives the complementary error function. Their values at $b$ and at infinity give the required boundary data.

For the upward-moving fresh-water/ice boundary, the Stefan condition is $\rho L\dot{a}=k{T}_{i,z}$: negative $\dot{a}$ means fresh water freezes. At the lower boundary the signed heat balance isSalt cannot enter the pure ice. With no bulk liquid motion its interfacial conservation law isThe derivatives and interface velocities areDefine $F(x)=\sqrt{\pi }x{e}^{x^2}\mathrm{erfc}x$. Substitution into all three balances givesTogether with $T_b=T_m-m{C}_b$, these determine the interfacial constants. In particular $C_b=C_0/[1-F({\lambda }_b)]$.

The physical branch has ${\lambda }_a<{\lambda }_b<0$. Since $F$ is negative at negative arguments, the salt relation gives $0<C_b<C_0$: lower-surface melting dilutes the brine. Combining the two heat balances eliminates the ice gradient and givesThus $T_m>T_b>T_{\infty }>T_L(C_0)$. The interface is at local liquidus equilibrium with diluted brine, while the undiluted far-field brine remains above its own liquidus. The similarity model describes a layer starting with zero thickness, or the long-time limit when an initial finite thickness no longer sets the dominant scale.

For the small-diffusivity-ratio limit, put $A=ϵ{\lambda }_a$ and $\Delta =T_m-T_{\infty }$, keeping $S=H/\Delta$ and the imposed concentrations fixed. Seek $A=O(1)$ and ${\lambda }_b=O(1)$. Then $F(ϵ{\lambda }_b)\sim \sqrt{\pi }ϵ{\lambda }_b$ and ${\lambda }_b-{\lambda }_a\sim -{\lambda }_a$. The upper heat balance givesIn the lower balance the two terms of order ${ϵ}^{-1}$ must cancel, givingConsequentlyand $T_b-T_{\infty }\sim -\sqrt{\pi }HA$. Adding the two temperature differences yields $2A^2-\sqrt{\pi }A=1/S$. The negative root isThe resulting $C_b=(T_m-T_b)/m$ lies strictly between zero and $C_0$. The salt relation therefore determines a finite negative ${\lambda }_b$, confirming its $O(1)$ scaling. The top interface travels on the thermal scale $\sqrt{\kappa t}$, whereas the lower melting interface travels on the shorter solutal scale $\sqrt{Dt}$. This is the brine-ice sandwich with disparate diffusion scales; $S$ is the latent-to-sensible heat ratio, the reciprocal of the usual sensible-to-latent Stefan number convention.

The sketches show constant fresh-water temperature, a declining linear ice temperature, and a much broader thermal than solutal boundary layer in the brine. Concentration is zero in freshwater and pure ice, jumps to $C_b$ across the lower interface, and rises to $C_0$. Accordingly the local liquidus is $T_m$ on the salt-free side, jumps to $T_b$ and falls rapidly to $T_L(C_0)$. In the phase diagram, the brine starts on the liquidus at $(C_b,T_b)$ and ends above it at $(C_0,T_{\infty })$; the pure ice has concentration zero, and the interface tie-line joins it to the diluted brine.

Physically, the cold saline liquid permits freshwater to freeze above, but salt lowers the liquidus and allows the existing ice to melt into brine below even though that brine is colder than $T_m$. Dilution raises the lower interfacial freezing temperature. Heat released by upper freezing supplies the lower melting and the conductive loss to the colder deep brine. With $|{\lambda }_a|\gg |{\lambda }_b|$, upper freezing exceeds lower melting and the ice layer thickens as $b-a\propto \sqrt{t}$.